The surface energy of a drop is E. If one thousand such drops coalesce to form a large drop then its surface energy is `E_1`. The ratio is `E_1` . The ratio `(E )/(E_1) ` is

Options

A. `1/100`
B. `1000/1`
C. `1/10`
D. `10/1`

Explanation

To solve the problem, we need to find the ratio of the surface energy of a small drop (E) to the surface energy of a large drop (E1) formed by the coalescence of 1000 small drops.

### Step-by-Step Solution:

1. **Understand the Surface Energy Formula**:
The surface energy (E) of a drop can be expressed as:
\[ E = \text{Surface Tension} \times \text{Area} \]
For a small drop of radius \( r \):
\[ E = S \times 4\pi r^2 \]
where \( S \) is the surface tension.

2. **Volume Conservation**:
When 1000 small drops coalesce to form a large drop, the volume of the large drop must equal the total volume of the 1000 small drops. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, for 1000 small drops:
\[ V_{\text{small}} = 1000 \times \frac{4}{3} \pi r^3 \]
For the large drop of radius \( R \):
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
Setting these volumes equal gives:
\[ \frac{4}{3} \pi R^3 = 1000 \times \frac{4}{3} \pi r^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ R^3 = 1000 r^3 \]
Taking the cube root:
\[ R = 10r \]

3. **Calculate Surface Energy of the Large Drop**:
Now, we can calculate the surface energy \( E_1 \) of the large drop:
\[ E_1 = S \times 4\pi R^2 \]
Substituting \( R = 10r \):
\[ E_1 = S \times 4\pi (10r)^2 = S \times 4\pi \times 100r^2 = 100 S \times 4\pi r^2 \]
Since \( E = S \times 4\pi r^2 \), we can express \( E_1 \) in terms of \( E \):
\[ E_1 = 100E \]

4. **Find the Ratio \( \frac{E}{E_1} \)**:
Now we can find the ratio:
\[ \frac{E}{E_1} = \frac{E}{100E} = \frac{1}{100} \]

### Final Answer:
The ratio \( \frac{E}{E_1} \) is \( \frac{1}{100} \).